Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $q \neq 0$. $n = \dfrac{5q^2 + 15q}{2q} \times \dfrac{-3}{4q + 12} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ (5q^2 + 15q) \times -3 } { 2q \times (4q + 12) } $ $ n = \dfrac {-3 \times 5q(q + 3)} {2q \times 4(q + 3)} $ $ n = \dfrac{-15q(q + 3)}{8q(q + 3)} $ We can cancel the $q + 3$ so long as $q + 3 \neq 0$ Therefore $q \neq -3$ $n = \dfrac{-15q \cancel{(q + 3})}{8q \cancel{(q + 3)}} = -\dfrac{15q}{8q} = -\dfrac{15}{8} $